What are the solutions to the equation t^2+8t=-12 ?

The solutions to the equation t^2+8t=-12 are t=-2,t=-6

Find the zeros of t^2+8t=-12

The zeros of t^2+8t=-12 are t=-2,t=-6

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t^{2} + 8 t = - 12

t = - 2, t = - 6

Solution steps

t^{2} + 8 t = - 12

Move

12

to the left side

t^{2} + 8 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 12 = 4

t_{1, 2} = \frac{- 8 \pm 4}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 8 + 4}{2 \cdot 1}, t_{2} = \frac{- 8 - 4}{2 \cdot 1}

t = \frac{- 8 + 4}{2 \cdot 1} : - 2

t = \frac{- 8 - 4}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

t = - 2, t = - 6

- 2 + 3 (0) \geq \frac{1}{4}

2 (x + 5) - x^{2} - 5 x = 0

\frac{( 2 a ^{- 1} ) ^{2}}{( 2 a ^{- 1} ) ^{- 2}}

61

2 x - 3 = 15

What are the solutions to the equation t^2+8t=-12 ?
The solutions to the equation t^2+8t=-12 are t=-2,t=-6
Find the zeros of t^2+8t=-12
The zeros of t^2+8t=-12 are t=-2,t=-6